The area enclosed by the curves $x^{2} = y, y = x + 2$ and x-axis is

\frac{1}{2}

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\frac{1}{3}

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\frac{1}{6}

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\frac{5}{6}

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Solution

The correct option is D $\frac{5}{6}$
Required Area is given by
$\int_{- 2}^{- 1} (x + 2) d x + \int_{- 1}^{0} x^{2} d x$
$= \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ .

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Similar questions

Find the area enclosed by the curve $y = x^{2}$ and the x-axis for interval x = 1 to x = 3.

y = 1 + x^{2}

y = 5

y = x^{2} + 1

y = 2 x + 1

The area enclosed by the curves y = |sin x|, x axis and |x| = $π$ is (in sq units)

| y + x | \leq 1, | y - x | \leq 1

2 x^{2} + 2 y^{2} = 1

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